The equation of a circle $C$ is $x^2+y^2-6x-16y+69 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Answer: To find the equation in standard form, complete the square. $(x^2-6x) + (y^2-16y) = -69$ $(x^2-6x+9) + (y^2-16y+64) = -69 + 9 + 64$ $(x-3)^{2} + (y-8)^{2} = 4 = 2^2$ Thus, $(h, k) = (3, 8)$ and $r = 2$.